/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Headway on four-lane highways: W... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 21

Headway on four-lane highways: When traffic is flowing on a highway, the headway is the average time between vehicles. On four-lane highways, the probability \(P\) that the headway is at least \(t\) seconds is given to a good degree of accuracy \({ }^{6}\) by $$ P=e^{-q t}, $$ a. On a four-lane highway carrying an average of 500 vehicles per hour in one direction, what is the probability that the headway is at least 15 seconds? (Note: 500 vehicles per hour is \(\frac{500}{3600}=0.14\) vehicle per second.) b. On a four-lane highway carrying an average of 500 vehicles per hour, what is the decay factor for the probability that headways are at least \(t\) seconds? Reminder: An important law of exponents tells us that \(a^{b c}=\left(a^{b}\right)^{c}\). where \(q\) is the average number of vehicles per second traveling one way on the highway.

Short Answer

Expert verified
a. Probability is approximately 0.1225. b. The decay factor is 0.14.

Step by step solution

01

Understanding the Problem

The headway on a highway is the average time between vehicles. We need to calculate two things using the headway probability function given by \( P = e^{-qt} \), where \( q \) is the average number of vehicles per second, and \( t \) is the time in seconds.
02

Convert Average Vehicles Per Hour to Per Second

Given that the highway has an average of 500 vehicles per hour traveling in one direction, convert this to vehicles per second by dividing by the number of seconds in an hour: \( q = \frac{500}{3600} \approx 0.14 \) vehicles per second.
03

Calculate Probability for Part a

To find the probability that the headway \( t \) is at least 15 seconds, plug \( q = 0.14 \) and \( t = 15 \) into the formula: \( P = e^{-0.14 \times 15} \). Simplify the exponent: \( -0.14 \times 15 = -2.1 \). Therefore, \( P = e^{-2.1} \). Using a calculator, \( e^{-2.1} \approx 0.1225 \).
04

Determine Decay Factor for Part b

The decay factor in the formula \( P = e^{-qt} \) is \( q \), which is the rate at which vehicles pass a point per second. Since we already calculated \( q = 0.14 \) vehicles per second, this is the decay factor.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Decay
The concept of exponential decay is crucial in various fields, including traffic flow analysis. It's a mathematical function that describes how a quantity decreases rapidly at first and then more slowly over time. In the context of traffic on a four-lane highway, this is described by the formula \( P = e^{-qt} \). Here, \( P \) represents the probability that the headway, or the interval between vehicles, meets or exceeds a certain time \( t \).
The term \( e^{-qt} \) involves the natural base \( e \), which is approximately 2.71828. When multiplied by a negative exponent, it reflects a form of decay 鈥� specifically, the chances decline as the headway time increases.
  • \( P \) is the probability that cars are spaced at least \( t \) seconds apart.
  • \( q \) is the decay constant, representing the rate of vehicle passage per second.
  • \( t \) is the headway time in seconds.
This exponential decay applies not only to headway but to any process where something decreases quickly then steadies. Understanding this pattern is key in analyzing how likely large headways are on busy highways.
Traffic Flow Analysis
Traffic flow analysis is a method used to understand how vehicles move along roadways. It involves the study of traffic patterns, speeds, and densities to optimize the movement of vehicles and reduce congestion. In our problem, traffic flow is assessed on a four-lane highway where headway probability helps to gauge traffic conditions.
A key aspect of analyzing traffic flow is understanding headway, which is the average time interval between vehicles. By calculating the headway, transportation planners can estimate the level of service on the highway. High headway values suggest lighter traffic, while lower values signify more congestion.
  • Analysis helps in planning road expansions or adjustments.
  • It's also essential for setting traffic lights and managing peak-hour traffic.
  • Applications include designing routes and predicting future traffic patterns.
By analyzing these aspects, planners aim to improve the efficiency of the road network, minimize travel times, and enhance overall road safety.
Rate of Vehicles
The rate of vehicles refers to the average number of vehicles passing a given point per unit of time, often expressed in vehicles per second or per hour. In this exercise, 500 vehicles per hour equates to a vehicle passage rate of approximately 0.14 vehicles per second (\[ q = \frac{500}{3600} \approx 0.14 \]). This number is pivotal in calculating the headway probability and understanding the decay factor.
The vehicle rate directly influences the headway since a higher rate suggests more frequent vehicle passage, reducing average headway. On highways, this rate helps determine road capacity and inform decisions regarding traffic management.
  • Higher rates (large \( q \)) result in shorter time intervals between cars.
  • Lower rates suggest less congestion and longer headways.
  • This rate provides insights crucial for urban transportation planning.
Overall, understanding the rate of vehicles is integral in adjusting infrastructure, optimizing travel times, and predicting traffic conditions effectively. This is central to both micro-level traffic studies and broader transportation policy developments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Grains of wheat on a chess board: A children's fairy tale tells of a clever elf who extracted from a king the promise to give him one grain of wheat on a chess board square today, two grains on an adjacent square tomorrow, four grains on an adjacent square the next day, and so on, doubling the number of grains each day until all 64 squares of the chess board were used. How many grains of wheat did the hapless king contract to place on the 64th square? There are about \(1.1\) million grains of wheat in a bushel. Assume that a bushel of wheat sells for \(\$ 4.25\). What was the value of the wheat on the 64th square?

$$ \begin{array}{|c|c|c|c|} \hline \text { Year } & \text { Wolves } & \text { Year } & \text { Wolves } \\\ \hline 1985 & 15 & 1993 & 40 \\ \hline 1986 & 16 & 1994 & 57 \\ \hline 1987 & 18 & 1995 & 83 \\ \hline 1988 & 28 & 1996 & 99 \\ \hline 1989 & 31 & 1997 & 145 \\ \hline 1990 & 34 & 1998 & 178 \\ \hline 1991 & 40 & 1999 & 197 \\ \hline 1992 & 45 & 2000 & 266 \\ \hline \end{array} $$ Gray wolves in Wisconsin: Gray wolves were among the first mammals protected under the Endangered Species Act in the \(1970 \mathrm{~s}\). Wolves recolonized in Wisconsin beginning in 1980 . Their population has grown reliably since 1985 as follows: \({ }^{21}\) a. Explain why an exponential model may be appropriate. b. Are these data exactly exponential? Explain. c. Find an exponential model for these data. d. Plot the data and the exponential model. e. Comment on your graph in part d. Which data points are below or above the number predicted by the exponential model?

. The half-life of U239: Uranium 239 is an unstable isotope of uranium that decays rapidly. In order to determine the rate of decay, 1 gram of U239 was placed in a container, and the amount remaining was measured at 1-minute intervals and recorded in the table below $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Time } \\ \text { in minutes } \end{array} & \begin{array}{c} \text { Grams } \\ \text { remaining } \end{array} \\ \hline 0 & 1 \\ \hline 1 & 0.971 \\ \hline 2 & 0.943 \\ \hline 3 & 0.916 \\ \hline 4 & 0.889 \\ \hline 5 & 0.863 \\ \hline \end{array} $$ a. Show that these are exponential data and find an exponential model. (For this problem, round all your answers to three decimal places.) b. What is the percentage decay rate each minute? What does this number mean in practical terms? c. Use functional notation to express the amount remaining after 10 minutes and then calculate that value. d. What is the half-life of U239?

Nearly linear or exponential data: One of the two tables below shows data that are better approximated with a linear function, and the other shows data that are better approximated with an exponential function. Make plots to identify which is which, and then use the appropriate regression to find models for both. $$ \begin{aligned} &\begin{array}{|c|c|} \hline t & f(t) \\ \hline 1 & 3.62 \\ \hline 2 & 23.01 \\ \hline 3 & 44.26 \\ \hline 4 & 62.17 \\ \hline 5 & 83.25 \\ \hline \end{array}\\\ &\begin{array}{|c|c|} \hline t & g(t) \\ \hline 1 & 3.62 \\ \hline 2 & 5.63 \\ \hline 3 & 8.83 \\ \hline 4 & 13.62 \\ \hline 5 & 21.22 \\ \hline \end{array} \end{aligned} $$

APR and APY: Recall that financial institutions sometimes report the annual interest rate that they offer on investments as the APR, often called the nominal interest rate. To indicate how an investment will actually grow, they advertise the annual percentage yield, or APY. \({ }^{7}\) In mathematical terms, this is the yearly percentage growth rate for the exponential function that models the account balance. In this exercise and the next, we study the relationship between the APR and the APY. We assume that the APR is \(10 \%\), or \(0.1\) as a decimal. To determine the APY when we know the APR, we need to know how often interest is compounded. For example, suppose for the moment that interest is compounded twice a year. Then to say that the APR is \(10 \%\) means that in half a year, the balance grows by \(\frac{10}{2} \%\) (or \(5 \%\) ). In other words, the \(\frac{1}{2}\) year percentage growth rate is \(\frac{0.1}{2}\) (as a decimal). Thus the \(\frac{1}{2}\)-year growth factor is \(1+\frac{0.1}{2}\). To find the yearly growth factor, we need to perform a unit conversion: One year is 2 half-year periods, so the yearly growth factor is \(\left(1+\frac{0.1}{2}\right)^{2}\), or \(1.1025\). a. What is the yearly growth factor if interest is compounded four times a year? b. Assume that interest is compounded \(n\) times each year. Explain why the formula for the yearly growth factor is $$ \left(1+\frac{0.1}{n}\right)^{n} $$ c. What is the yearly growth factor if interest is compounded daily? Give your answer to four decimal places.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.